Question

Initially (Chintu + disc) are at rest.

Suddenly chintu starts moving with $v_0$ wrt ground.

$\omega = ?$ disc

Answer 1

$\frac{2 m_c v_0}{M_d R}$

Answer 2

The system (Chintu + disc) is initially at rest, so its total angular momentum is zero. As no external torque acts on the system, angular momentum is conserved. When Chintu moves with velocity $v_0$ (wrt ground) at radius $R$, his angular momentum is $m_c v_0 R$. To conserve total angular momentum, the disc must rotate in the opposite direction, generating an angular momentum $I_d \omega = \frac{1}{2} M_d R^2 \omega$. Equating the magnitudes of these opposing angular momenta ($m_c v_0 R = \frac{1}{2} M_d R^2 \omega$) yields $\omega = \frac{2 m_c v_0}{M_d R}$.

Answer:

The angular velocity of the disc ($\omega$) is $\frac{2 m_c v_0}{M_d R}$.